experiment 1 distance traveled by a projectile

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• Exploring Projectile Motion: A Comprehensive Guide to Understanding and Conducting Experiments
• Physics experiments

Welcome to our comprehensive guide on exploring and conducting experiments on projectile motion . This article is a part of our Silo on physics and classical mechanics experiments, and it aims to provide you with a thorough understanding of this fundamental concept in physics. Whether you are a student, a teacher, or simply someone interested in the mechanics of motion, this article is the perfect resource for you. So buckle up and get ready to dive into the world of projectile motion experiments! In this article, we will cover everything you need to know about projectile motion experiments, from the basic principles and equations to the step-by-step process of conducting your own experiment.

We will also discuss the real-life applications of projectile motion and how it relates to other concepts in physics. By the end of this article, you will have a solid understanding of projectile motion and be able to confidently conduct your own experiments. So let's get started!Welcome to our comprehensive guide on understanding and conducting experiments for projectile motion! Whether you are a student studying classical mechanics or a curious individual interested in physics experiments, this article is for you. We will delve into the fundamentals of projectile motion, its applications, and how to conduct your own experiments to better understand this phenomenon.

So, let's get started and explore the fascinating world of projectile motion!To begin, let's define projectile motion . It is the motion of an object that is projected into the air at an angle and follows a curved path due to the force of gravity. This type of motion can be observed in everyday life, from a ball thrown in a game of catch to a rocket launching into space. The key to understanding projectile motion lies in knowing its fundamental principles, such as velocity, acceleration, and trajectory.

These concepts are described by various formulas that we will cover in detail. For instance, the formula for calculating the horizontal distance traveled by a projectile is d = v * t * cosθ, where d is distance, v is initial velocity, t is time, and θ is launch angle. These formulas may seem intimidating at first, but with practice and examples, you'll become proficient in using them to solve problems. It's important to note that understanding projectile motion is not just about memorizing formulas; it's about grasping the underlying principles and applying them to real-world scenarios. To begin, let's define projectile motion .

The key to understanding projectile motion lies in knowing its fundamental principles, such as velocity, acceleration, and trajectory. These concepts are described by various formulas that we will cover in detail. For instance, the formula for calculating the horizontal distance traveled by a projectile is d = v * t * cosθ , where d is distance, v is initial velocity, t is time, and θ is launch angle. These formulas may seem intimidating at first, but with practice and examples, you'll become proficient in using them to solve problems. It's important to note that understanding projectile motion is not just about memorizing formulas; it's about grasping the underlying principles and applying them to real-world scenarios. Are you fascinated by the laws of motion and the forces that govern our physical world? Do you enjoy hands-on experiments and solving complex problems? Then you've come to the right place! In this article, we will dive into the world of projectile motion and explore its concepts, formulas, tutorials, and resources.

Solving Problems

Conducting experiments.

The best way to understand a concept is to see it in action. Using simple materials like a ball, a ruler, and a stopwatch, you can set up an experiment to observe the motion of a projectile. Start by launching the ball at different angles and measuring the distance traveled. Then, vary the initial velocity and record the results.

Finding Tutorials and Resources

Pursuing a career in physics.

Whether you're a student, a researcher, or simply curious about physics, we hope this article has provided valuable insights and resources to further your exploration of projectile motion. In conclusion, projectile motion is a fascinating concept that has real-world applications and is integral to understanding the laws of motion. Whether you're a student, a researcher, or simply curious about physics, we hope this article has provided valuable insights and resources to further your exploration of projectile motion.

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Experiment 1 Distance Traveled by a Projectile In this experiment you will use kinematic equations to predict the range of a projectile set in motion To do this you will roll marbles down a ramp and off a table to observe vertical and horizontal motion Materials Sheet of Carbon Paper 1 Fishing Sinker Masking Tape 1 Marble Monofilament Line 1 Protractor 1 Ramp Sheet of Printer Paper Tape Measure Pencil Table You Must Provide Note You will need to construct the ramp provided in your lab kit prior to beginning the experiment To do this complete the following steps Ramp Set Up Figure 4 1 Separate the two pieces one long and narrow piece to provide the ramp and one wider piece to provide the base Figure 4 Ramp set up diagram 2 Fold the wider section along the perforations to form a triangular stand 3 Insert the tab through the slot to construct a triangular stand Figure 4 Part 2 4 Insert the tab on long narrow piece into one of three slots on the triangular stand Different slots correspond to different inclines Procedure 2014 eScience Labs LLC All Rights Reserved 1 Find a table upon which to perform the experiment Place the ramp so that its bottom edge is positioned at the edge of the table You will be rolling marbles down the ramp and off the table in this experiment 2 Use a protractor to measure the incline of your ramp Record the incline in Table 1 3 Use a pencil to mark three different locations on the ramp at which you will release the marble This will ensure the marble achieves the same velocity with each trial Hint Use locations near the top middle and bottom of the ramp 4 Create a plumb line by attaching the fishing sinker to the monofilament line 5 Hold the string to the edge of the table and use a piece of masking tape to mark the spot at which the weight touches the ground Note The length of the plumb line will help you measure the exact distance from the edge of the ramp to the position where the marble lands 6 Begin the experiment by releasing the marble from the first position you marked on the ramp in Step 3 In other words release the marble from the highest position which you marked on the ramp 7 Carefully observe where the marble hits the ground and place a piece of white printer paper at that location Secure the paper to the ground with a small piece of masking tape Make sure the paper can moved when the different ramp positions are tested Try to center the printer paper over the spot where the marble hit the floor 8 Set the carbon paper on the printer paper so that the light side faces up When the marble hits the carbon paper it will leave a mark on the printer paper 9 Place the marble at the same drop mark you just tested and release it 10 Observe and measure the distance traveled to the first mark made on the printer paper using the tape measure The mark may be faint but it will be visible Record this value in Table 1 11 Once you have recorded the distance in Table 1 put an X over the mark you just measured so you do not reuse it 12 Repeat Steps 9 10 three more times and record your data in Table 1 13 Repeat Steps 6 12 for the remaining two ramp distances you marked in Step 2 Record you results for the second ramp distance in Table 2 and the third ramp distance in Table 3 14 Save the printer paper The unused side will be used in the next experiment Table 1 Range and Velocity of Projectile at Ramp Distance 1 Ramp Incline degrees 2014 eScience Labs LLC All Rights Reserved Ramp Distance m Trial Measured Distance m 1 2 3 4 Average Table 2 Range and Velocity of Projectile at Ramp Distance 2 Ramp Distance m Trial Measured Distance m 1 2 3 4 Average Table 3 Range and Velocity of Projectile at Ramp Distance 3 Ramp Distance m Trial Measured Distance m 1 2 3 4 Average 2014 eScience Labs LLC All Rights Reserved Post Lab Questions 1 Use your predictions of velocity and range from the Pre Lab Questions and the data recorded from your experiment to complete Table 4 Table 4 Velocity and Range Data for all Ramp Distances Ramp Distance m Calculated velocity m s Predicted Range m Average Actual Range m Percent Error 2 How do your predictions compare to the observed data Explain at least two reasons for the differences 3 If you were to fire a paintball pellet horizontally and at the same time drop the same type of paintball pellet you fired from the paintball gun which pellet would hit the ground first and why is this so 4 Suppose you altered your existing ramp so that the marbles had twice their initial velocity right before leaving the ramp How would this change the total distance traveled and the time that the marbles were in the air 5 Describe the acceleration of the marbles after it leaves the ramp Use kinematic equations to support your discussion 2014 eScience Labs LLC All Rights Reserved

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Question: EXPERIMENT 1: DISTANCE TRAVELED BY A PROJECTILE Data Sheet Table 1. Range and Velocity of Projectile at Ramp Distance 1 Table 2. Range and Velocity of Projectile at Ramp Distance 2\r\nTable 3. Range and Velocity of Projectile at Ramp Distance 3 Post-Lab Questions 1. Use your predictions of velocity and Range from the Pre-Lab Questions and the data recorded

1. Calculate Initial Velocity ( ( v initial ) ) :

The initial velocity of the projectile can be calculated using the kin...

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4.3 Projectile Motion

Learning objectives.

By the end of this section, you will be able to:

• Use one-dimensional motion in perpendicular directions to analyze projectile motion.
• Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
• Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
• Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure 4.11 illustrates the notation for displacement, where we define s → s → to be the total displacement, and x → x → and y → y → are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y .

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y -axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

Because gravity is vertical, a x = 0 . a x = 0 . If a x = 0 , a x = 0 , this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0 x . v x = v 0 x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration . The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = − g , a x = 0 : a y = − g , a x = 0 :

Horizontal Motion

Vertical Motion

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy

Projectile motion.

• Resolve the motion into horizontal and vertical components along the x - and y -axes. The magnitudes of the components of displacement s → s → along these axes are x and y. The magnitudes of the components of velocity v → v → are v x = v cos θ and v y = v sin θ , v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Figure 4.12 .
• Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
• Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
• Recombine quantities in the horizontal and vertical directions to find the total displacement s → s → and velocity v → . v → . Solve for the magnitude and direction of the displacement and velocity using s = x 2 + y 2 , Φ = tan −1 ( y / x ) , v = v x 2 + v y 2 , s = x 2 + y 2 , Φ = tan −1 ( y / x ) , v = v x 2 + v y 2 , where Φ is the direction of the displacement s → . s → .

Example 4.7

A fireworks projectile explodes high and away.

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , v 0 y , the component of the initial velocity in the y direction. It is given by v 0 y = v 0 sin θ 0 , v 0 y = v 0 sin θ 0 , where v 0 v 0 is the initial velocity of 70.0 m/s and θ 0 = 75 ° θ 0 = 75 ° is the initial angle. Thus,

Thus, we have

Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.

(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0 y − g t . v y = v 0 y − g t . Because v y = 0 v y = 0 at the apex, this equation reduces to simply

This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + 1 2 ( v 0 y + v y ) t . y = y 0 + 1 2 ( v 0 y + v y ) t . This is left for you as an exercise to complete.

(c) Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t , x = x 0 + v x t , where x 0 x 0 is equal to zero. Thus,

where v x v x is the x -component of the velocity, which is given by

Time t for both motions is the same, so x is

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.

(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point:

Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure 4.11 , which shows the curvature of the trajectory toward the ground level.

When solving Example 4.7 (a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h . Then,

This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Check Your Understanding 4.3

A rock is thrown horizontally off a cliff 100.0 m 100.0 m high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

Example 4.8

Calculating projectile motion: tennis player.

If we take the initial position y 0 y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

Substituting into Equation 4.22 for y gives us

Rearranging terms gives a quadratic equation in t :

Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator:

The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.

(b) We can find the final horizontal and vertical velocities v x v x and v y v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector v → v → and the angle θ θ it makes with the horizontal. Since v x v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore,

The final vertical velocity is given by Equation 4.21 :

Since v 0 y v 0 y was found in part (a) to be 21.2 m/s, we have

The magnitude of the final velocity v → v → is

The direction θ v θ v is found using the inverse tangent:

Significance

Time of flight, trajectory, and range.

Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flight

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

Factoring, we have

Solving for t gives us

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation 4.24 does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g . Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y ( x ). We take x 0 = y 0 = 0 x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

Substituting the expression for t into the equation for the position y = ( v 0 sin θ 0 ) t − 1 2 g t 2 y = ( v 0 sin θ 0 ) t − 1 2 g t 2 gives

Rearranging terms, we have

This trajectory equation is of the form y = a x + b x 2 , y = a x + b x 2 , which is an equation of a parabola with coefficients

From the trajectory equation we can also find the range , or the horizontal distance traveled by the projectile. Factoring Equation 4.25 , we have

The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

corresponding to the impact point. Using the trigonometric identity 2 sin θ cos θ = sin 2 θ 2 sin θ cos θ = sin 2 θ and setting x = R for range, we find

Note particularly that Equation 4.26 is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed v 0 v 0 and sin 2 θ 0 sin 2 θ 0 , and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor sin 2 θ 0 sin 2 θ 0 that the range is maximum at 45 ° . 45 ° . These results are shown in Figure 4.15 . In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45 ° . 45 ° . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to 90 ° . 90 ° . The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Example 4.9

Comparing golf shots.

(a) What is the initial speed of the ball at the second hole?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

(b) R = v 0 2 sin 2 θ 0 g ⇒ v 0 = R g sin 2 θ 0 = 90.0 m ( 9.8 m / s 2 ) sin ( 2 ( 70 ° ) ) = 37.0 m / s R = v 0 2 sin 2 θ 0 g ⇒ v 0 = R g sin 2 θ 0 = 90.0 m ( 9.8 m / s 2 ) sin ( 2 ( 70 ° ) ) = 37.0 m / s

(c) y = x [ tan θ 0 − g 2 ( v 0 cos θ 0 ) 2 x ] Second hole: y = x [ tan 30 ° − 9.8 m / s 2 2 [ ( 31.9 m / s)( cos 30 ° ) ] 2 x ] = 0.58 x − 0.0064 x 2 Fourth hole: y = x [ tan 70 ° − 9.8 m / s 2 2 [ ( 37.0 m / s)( cos 70 ° ) ] 2 x ] = 2.75 x − 0.0306 x 2 y = x [ tan θ 0 − g 2 ( v 0 cos θ 0 ) 2 x ] Second hole: y = x [ tan 30 ° − 9.8 m / s 2 2 [ ( 31.9 m / s)( cos 30 ° ) ] 2 x ] = 0.58 x − 0.0064 x 2 Fourth hole: y = x [ tan 70 ° − 9.8 m / s 2 2 [ ( 37.0 m / s)( cos 70 ° ) ] 2 x ] = 2.75 x − 0.0306 x 2

(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure 4.16 .

Check Your Understanding 4.4

If the two golf shots in Example 4.9 were launched at the same speed, which shot would have the greatest range?

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure 4.17 , which is based on a drawing in Newton’s Principia. If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation .

Interactive

At PhET Explorations: Projectile Motion , learn about projectile motion in terms of the launch angle and initial velocity.

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• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
• Publisher/website: OpenStax
• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion

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